So if you have an empty memory card compared to one which is full with say 64GB of data (or any other large amount). If you weighed them both on an incredibly sensitive scales, would there be a difference in the weight?
Essentially I’m asking does data/computer memory have any mass?
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As other comments got to, this depends on your storage technology.
Since flash memory and other “solid state” memory is holding electrons in capacitors to count as a “1”, by filling that 64gb flash drive with 1’s you’re adding the weight of ~ 64 billion electrons to it (if it started at all 0’s, and it is storing a singular electron and not just some level of charge, which I don’t know is the case). However, the weight of 64 billion electrons is 9.05 x 10^-28 grams/electron * 6.4×10^9 electrons for a combined added mass of 5.792 x 10^-18 grams. Or 0.000000000000000005792 grams.
Some rough back-of-the-envelope math tells me that we will need solid state hardware capable of holding roughly tens of thousands of yottabytes (a yottabyte is 2^80 bytes, which is in the neighborhood of 10^24; a gibibyte is 2^30, a gigabyte is 10^9, most storage measures in gibibytes even if they say gigabytes because gibibyte is a more recent term that hasn’t entered widespread usage yet to resolve the ambiguity around whether a gigabyte was 10^9 bytes or 2^30 bites)
However, to this question:
>Essentially I’m asking does data/computer memory have any mass?
The inherent answer is “no” – because historically data has been (and in many places still is) stored on magnetic media that relied on the orientation of a magnetic particle rather than the presence/absence of a particle to indicate a 1 or a 0. Data is an abstraction that we use to build immensely complex things, and inherently an abstraction has no mass, unless the means by which we notate it, encode it, store it, etc – has mass that would change depending on the state of the data being a 1 or 0.
**TL;DR: Yes, but actually no.**
PRE-EMPTIVE edit: I just realized that 64 gibi*bytes* is 8 times as many *bits* which is what the electrons would represent, so multiply by 8 to get a more precise answer, but the short version is that no amount of storage we are likely to get any time soon is going to increase the weight of the device by an appreciable margin by storing all 1’s as opposed to all 0’s.
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